Little orthogonality theorem

The little orthogonality theorem consists of two relations that are reduced forms of the mathematical expression for the great orthogonality theorem.

1^st little orthogonality relation

The 1^st relation is derived from eq21 by letting $j=i$ and $j'=i'$ , resulting in

$\sum_{\alpha=1}^{\vert G\vert}\Gamma_k^*(\alpha)_{ii}\Gamma_{k'}(\alpha)_{i'i'}=\frac{\vert G\vert}{d}\delta_{ii'}\delta_{ii'}\delta_{kk'}$

Since $(I)_{ii'}(I)_{ii'}=(I)_{ii'}$ , we have $\delta_{ii'}\delta_{ii'}=\delta_{ii'}$ and

$\sum_{\alpha=1}^{\vert G\vert}\Gamma_k^*(\alpha)_{ii}\Gamma_{k'}(\alpha)_{i'i'}=\frac{\vert G\vert}{d}\delta_{ii'}\delta_{kk'}$

$\sum_{i=1}^n\sum_{i'=1}^n\sum_{\alpha=1}^{\vert G\vert}\Gamma_k^*(\alpha)_{ii}\Gamma_{k'}(\alpha)_{i'i'}=\sum_{i=1}^n\sum_{i'=1}^n\frac{\vert G\vert}{d}\delta_{ii'}\delta_{kk'}$

$\sum_{\alpha=1}^{\vert G\vert}\sum_{i=1}^n\Gamma_k^*(\alpha)_{ii}\sum_{i'=1}^n\Gamma_{k'}(\alpha)_{i'i'}=\frac{\vert G\vert}{d}\delta_{kk'}\sum_{i=1}^n\sum_{i'=1}^n\delta_{ii'}$

$\sum_{\alpha=1}^{\vert G\vert}tr\[\Gamma_k^*(\alpha)\]tr\[\Gamma_{k'}(\alpha)\]=\frac{\vert G\vert}{d}\delta_{kk'}d=\vert G\vert\delta_{kk'}\;\;\;\;\;\;\;\;22$

Question

Show that $\sum_{i=1}^n\sum_{i'=1}^n\delta_{ii'}=d$ .

Answer

$\sum_{i=1}^n\sum_{i'=1}^n\delta_{ii'}=\sum_{i=1}^n(\delta_{i1}+\delta_{i2}+\cdots+\delta_{in})$

$=(\delta_{11}+\delta_{21}+\cdots+\delta_{n1})+(\delta_{12}+\delta_{22}+\cdots+\delta_{n2})+(\delta_{1n}+\delta_{2n}+\cdots+\delta_{nn})$

$=\delta_{11}+\delta_{22}+\cdots+\delta_{nn}=1+1+\cdots+1=d$

Since the traces of a matrices of the same class are the same, eq22 is equivalent to

$\sum_{\alpha=1}^{C}n_{\alpha}\chi_k^*(\alpha)\chi_{k'}(\alpha)=\vert G\vert\delta_{kk'}\;\;\;\;\;\;\;\;23$

where

1. $C$ is the number of classes.
2. $n_{\alpha}$ is the number of elements in the $\alpha$ -th class.
3. $\chi_k(\alpha)$ , called the character of a class, is the trace of a matrix belonging to the $\alpha$ -th class of the $k$ -th irreducible representation.

Eq23 is known as the 1^st little orthogonality relation.

Question

Determine whether the representations $\Gamma_1$ and $\Gamma_2$ of the $C_{3v}$ point group are reducible or irreducible.

$\begin{matrix}&E&2C_3&3\sigma_v\\\Gamma_1&1&1&1\\\Gamma_2&3&0&1\end{matrix}$

Answer

Since the 1^st little orthogonality relation is derived from the great orthogonality theorem, which pertains only to irreducible representations, characters of a representation that do not satisfy the relation belong to a reducible representation. Using eq23,

$\begin{matrix}\Gamma_1&1\cdot 1\cdot 1+2\cdot 1\cdot 1+3\cdot 1\cdot 1=6\cdot 1\\\Gamma_2&1\cdot 3\cdot 3+2\cdot 0\cdot 0+3\cdot 1\cdot 1\neq 6\cdot 1\end{matrix}$

Therefore, $\Gamma_1$ is an irreducible representation of the $C_{3v}$ point group, while $\Gamma_2$ is a reducible representation of the $C_{3v}$ point group.

Let’s rewrite eq23 as

$\sum_{\alpha=1}^{C}\sqrt{\frac{n_{\alpha}}{\vert G\vert}}\chi_k^*(\alpha)\sqrt{\frac{n_{\alpha}}{\vert G\vert}}\chi_{k'}(\alpha)=\delta_{kk'}\;\;\;\;\;\;\;\;24$

Eq24 has the form of the inner product of two vectors $\boldsymbol{\mathit{u}}$ and $\boldsymbol{\mathit{v}}$ in a $C$ -dimensional vector space, with components $\sqrt{\frac{n_{\alpha}}{\vert G\vert}}\chi_k^*(\alpha)$ and $\sqrt{\frac{n_{\alpha}}{\vert G\vert}}\chi_{k'}(\alpha)$ respectively. The components are functions of $k$ and the two vectors are orthogonal to each other when $k\neq k'$ . A $C$ -dimensional vector space is spanned by $C$ orthogonal vectors, each with $C$ components. Since the number of vectors and the number of components of each vector are denoted by $k$ and $\alpha$ respectively, the number of irreducible representations of a group is equal to the number of classes of that group. We say that the irreducible representations of a group form a complete set of basis vectors in the $C$ -dimensional vector space, with the components of each basis vector being $\sqrt{\frac{n_{\alpha}}{\vert G\vert}}\chi_{k'}(\alpha)$ .

2^nd little orthogonality relation

Consider the matrices $Q$ and $Q'$ with entries $Q_{ij}=\chi_i(j)$ and $Q'_{ij}=\frac{n_i}{\vert G\vert}\chi^*_j(i)$ respectively.

$Q=\begin{pmatrix}\chi_1(1)&\chi_1(2)&\cdots &\chi_1(j)\\\chi_2(1)&\chi_2(2)&\cdots &\chi_2(j)\\\vdots &\vdots &\ddots &\vdots\\\chi_i(1)&\chi_i(2)&\cdots &\chi_i(j)\end{pamatrix}\;\;\;Q'=\frac{1}{\vert G\vert}\begin{pmatrix}n_1\chi^*_1(1)&n_1\chi^*_2(1)&\cdots &n_1\chi^*_j(1)\\n_2\chi^*_1(2)&n_2\chi^*_2(2)&\cdots &n_2\chi^*_j(2)\\\vdots &\vdots &\ddots &\vdots\\n_i\chi^*_1(i)&n_i\chi^*_2(i)&\cdots &n_i\chi^*_j(i)\end{pamatrix}$

Both are square matrices because the number of irreducible representations of a group is equal to the number of classes of that group. Comparing with eq24, the entries of the matrix product $QQ'$ are given by:

$(QQ')_{pq}=\sum_{k=1}^C\chi_p(k)\frac{n_k}{\vert G\vert}\chi_q^*(k)=\delta_{pq}\;\;\;\;\;\;\;\;24a$

Since $Q$ and $Q'$ are square matrices and $QQ'=I$ , then $Q'Q=I$ , i.e.

$(Q'Q)_{pq}=\sum_{k=1}^C\frac{n_p}{\vert G\vert}\chi_k^*(p)\chi_k(q)=\delta_{pq}\;\;\;\;\;\;\;\;24b$

Eq24b is the 2^nd little orthogonality relation.

Question

Show that if $Q$ and $Q'$ are square matrices and $QQ'=I$ , then $Q'Q=I$ .

Answer

Using the determinant identities of $\vert AB\vert=\vert A\vert\vert B\vert$ and $\vert I\vert=1$ , we have $\vert QQ'\vert=\vert Q\vert\vert Q'\vert=\vert I\vert=1$ , which implies that $\vert Q\vert$ and $\vert Q'\vert$ are not zero and are therefore non-singular. So,